Exercises: Sine, Cosine and Tangent on a right triangle

Hint: Proceed backwards in pictures:

For this exercise, you need to be able to use the sine and cosine in the right triangle and the Pythagorean theorem.

Strategy

If you want to work out a strategy for the solution first, it is best to work backwards:

To calculate the length of $x$, for example, you need all the other distances in this triangle. You know the length of $\overline{BC}$. It is the same length as that of $\overline{AD}$, because they are opposite sides in a rectangle. That means you only have to determine $\overline{BE}$.

$\overline{BE}$ can be calculated by subtracting the length of $\overline{AB}$ from the long side $\overline{AE}$. You can calculate $\overline{AE}$ with the using the tangent in the triangle $\mathrm{\Delta }ADE$. To calculate $\overline{AB}$, for example, you can use the tangent in the triangle $\mathrm{\Delta }ACD$, because you know that $\overline{DC}$ and $\overline{AB}$, as opposite sides in the rectangle, have the same length.

Lösung

Nun kennst du das Vorgehen "von hinten" und kannst es in genau umgekehrter Reihenfolge verwenden, um auf die Länge der Strecke $x$ zu kommen:

Berechnung von $\overline{DC}$

Verwende den Tangens im Dreieck $\mathrm{\Delta }ACD$ mit dem dir bekannten Winkel $\measuredangle \mathrm{CAD}$ für die Berechnung von $\overline{DC}$:

Solution

Now you know the procedure "from behind" and can use it in exactly the opposite order to arrive at the length of $x$:

Calculation of $\overline{DC}$

Use the tangent in the triangle $\mathrm{\Delta }ACD$ with the known angle $\measuredangle \mathrm{CAD}$ to calculate $\overline{DC}$:

$$\tan (\measuredangle \mathrm{CAD})={\displaystyle \frac{\text{opposite cathetus}}{\text{adjacent cathetus}}=\frac{\overline{DC}}{\overline{AD}}}$$

Solve for $\overline{DC}$ by multiplying both sides with $\overline{AD}$.

$$\overline{DC}={\displaystyle \overline{AD}\cdot \tan (\measuredangle \mathrm{CAD})}$$

Plug in the values $\overline{\mathrm{AD}}=7\mathrm{m}$ and $\measuredangle \mathrm{CAD}={50}^{\circ }$ .

$$\overline{DC}=\overline{AB}={\displaystyle 7\cdot \tan ({50}^{\circ })\approx 8.34}$$

Computing $\overline{AE}$

Use the tangent in the triangle $\mathrm{\Delta }ADE$ to calculate $\overline{AE}$, since you know $\measuredangle \mathrm{ADE}={55}^{\circ }$ and $\overline{AD}=7$.

$$\tan (\measuredangle \mathrm{ADE})={\displaystyle \frac{\text{opposite cathetus}}{\text{adjacent cathetus}}=\frac{\overline{AD}}{\overline{AE}}}$$

Solve for $\overline{AE}$ by multiplying with $\overline{AE}$ and dividing by $\tan (\measuredangle \mathrm{ADE})$.

$$\overline{AE}={\displaystyle \frac{\overline{AD}}{\tan (\measuredangle \mathrm{ADE})}}$$

Plug in the values.

$$\overline{AE}={\displaystyle \frac{7}{\tan ({55}^{\circ })}\approx 10.00}$$

Computing $\overline{BE}$

Now you can compute $\overline{BE}$ :

$\overline{BE}=\overline{AE}-\overline{AD}=10.00-8.34=1.66m$

Computing $x$

Now you can calculate the length of $x$ using the Pythagorean theorem. Here $x$ is the hypotenuse.

$$x^2={\displaystyle {\overline{BE}}^2+{\overline{BC}}^2}$$

Solve for $x$ by taking the root.

$$x={\displaystyle \sqrt{{\overline{BE}}^2+{\overline{BC}}^2}}$$

Plug in the values. Remember that you can use $\overline{BC}=\overline{AD}$ because they are opposite sides in the rectangle.

$$x={\displaystyle \sqrt{{1.66}^2+7^2}\approx 7.19m}$$

The side $x$ is $7.19m$ long.

Here, as is very often the case, there is not only one possible solution. For example, you can also use the sine or cosine instead of the tangent with another intermediate step.